Algebra in the Punjab Police SI paper is not about theory. Every one of the 32 algebra questions asked since 2016 is a sentence you have to turn into an equation, or an ugly expression that collapses the moment you spot the identity hiding in it. That gives you only two skills to build. Translate: give the unknown a letter, write one equation per sentence, solve. Recognise: learn eight identities so well that (0.93³ − 0.7³)/(0.93² + 0.651 + 0.7²) is a two-second answer instead of a calculator job. This note takes both, part by part — identities, linear pairs, quadratics, ages, variation and the single logarithm question — with every SI question solved.
The eight identities
Nine of the 32 questions are one of these in disguise.
This single line, read backwards, is the SI examiner's favourite trick. When you see a fraction whose denominator is a² − ab + b² and whose numerator is a³ + b³, the whole thing is just 1/(a + b). The sign rule is worth saying aloud: same sign outside, opposite sign in the middle. a³ + b³ takes (a + b) outside and −ab in the middle; a³ − b³ takes (a − b) outside and +ab in the middle.
a² + b² = (a + b)² − 2ab and a² + b² = (a − b)² + 2ab. A question almost never gives you a and b separately; it gives you their sum and their product, or their difference and their product. These two lines convert what you were given into what was asked, with no solving at all. The cube version works the same way: a³ − b³ = (a − b)³ + 3ab(a − b).
Turning sentences into equations
The largest group: fourteen questions.
- Name the unknown
- One sentence, one equation
- Eliminate
- Answer what was asked
If (3/5) × (x ÷ 6/5 − 2 ÷ 3) 5 = 3y and y = 4(4 ÷ 2/5 × 2 − 2 ÷ 5/3) ÷ 4/5, then which of the following is a whole number? (SI 14 Jul 2023)
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Work out y first, strictly left to right inside the bracket. 4 ÷ 2/5 = 10, and 10 × 2 = 20. Next 2 ÷ 5/3 = 6/5 = 1.2, so the bracket is 20 − 1.2 = 18.8. Then 4 × 18.8 = 75.2, and 75.2 ÷ 4/5 = 75.2 × 5/4 = 94. So y = 94. Now x. The left side is (3/5) × (5x/6 − 2/3) × 5 = 3(5x/6 − 2/3), and this equals 3y = 282. So 5x/6 − 2/3 = 94, giving 5x/6 = 284/3 and x = 568/5 = 113.6. Test the options: x + (y + 3)/5 = 113.6 + 97/5 = 113.6 + 19.4 = 133, a whole number. The other three give 359.6, 132.8 and 132.6.
Quadratics without the formula
Sum and product beat substitution every time.
D = b² − 4ac. If D > 0 there are two distinct real roots; if D = 0 the two roots are equal; if D < 0 there is no real root. If D is also a perfect square and a, b, c are integers, the roots are rational — which is exactly the test to apply when a question says “the roots are integers”.
Ages
Four questions, one table.
The same puzzle about X, A, B, C, D and E appears in three different SI papers, each time asking for a different person. Solve it once, in this order: X = √25 = 5 · C = 12X = 60 · A = 10% of C = 6 · B = X + 2.8 = 7.8 · D × A = 10B ⇒ D = 78/6 = 13 · E: D + 2 = 2E ⇒ E = 7.5 Always start from the one sentence that gives a bare number — here “X is the square root of 25” — and chain outwards. In the 15 July 2022 version B is defined as 30% more than A instead, which changes B but not C, so the answer to “What is C's age?” is still 60.
There are six persons X, A, B, C, D and E. A's age is 10% of C's age. After 2 years, D's age will be 2 times E's age. Also, the product of D's age and A's age is 10 times B's age, and C's age is 12 times X's age. B's age is 2.8 years more than X's age, which is the square root of 25. What is A's age? (SI 20 Jun 2023)
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X = √25 = 5 and C = 12X = 60, so A = 10% of 60 = 6. The rest of the data (B = 7.8, D = 13) is not needed for this one.
There are six persons X, A, B, C, D and E. A's age is 10% of C's age. After 2 years, D's age will be 2 times E's age. Also, the product of D's age and A's age is 10 times B's age, and C's age is 12 times X's age. B's age is 2.8 years more than X's age, which is the square root of 25. What is D's age? (SI 29 Jul 2023)
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Same chain: X = 5, C = 60, A = 6, B = X + 2.8 = 7.8. Then D × A = 10B gives 6D = 78, so D = 13.
There are six people X, A, B, C, D, E. B’s age is 30% more than A’s age, which is 10% of C’s age. D’s age 2 years later will be 2 times the age of E. Also, the product of D’s age and A’s age is 10 times that B’s age and C’s age is 12 times that of X’s age. E’s age is 1.5 times that of X’s age, which is the square root of 25. What is C’s age? (SI 15 Jul 2023)
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This version defines B as 30% more than A, but the question only asks for C. X = √25 = 5 and C = 12X = 60 — one line, and the rest of the puzzle can be ignored. Read what is asked before you solve anything.
Variation and the one logarithm question
Small, but they come back.
Inverse variation is a constant product, not a constant ratio. Find k from the pair you are given, then put the new value in. Direct variation is the mirror image: y = kx, a constant ratio. The SI paper has asked the inverse-square version twice with different numbers, so learn the two-step drill — find k, then substitute — rather than the answers.
Given that q = 4 for p = 1 and p varies inversely as square of q. Which of the following options is equal to the value of p for q = 8? (SI 31 Jul 2023)
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Same drill: k = pq² = 1 × 16 = 16. At q = 8, p = 16 ÷ 64 = 1/4. Again q doubled and p fell to a quarter.
- Answering x when the question asked for x + y, for a sum of ages, or for half the capacity. Underline the last line of the stem before you start.
- Squaring surds or decimals one at a time. If you see a conjugate pair or a difference of squares, use the identity — it is always faster and always exact.
- Forgetting that a loose decimal in a denominator is usually ab. 0.651 is 0.93 × 0.7.
- In an ages question with two children, adding 6 to the total instead of 12. Each person ages separately.
- Treating inverse variation as a ratio. pq² = k is a product; doubling q divides p by four, not by two.
- a³ + b³ = (a + b)(a² − ab + b²)
- a³ − b³ = (a − b)(a² + ab + b²)
- a² + b² = (a + b)² − 2ab
- Roots: sum = −b/a, product = c/a
- D = b² − 4ac decides how many real roots
- Remainder of p(x) ÷ (x + 1) is p(−1)
- Inverse variation: pqⁿ = k
- 1/log_b a = log_a b
Same outside, opposite inside. a³ + b³ keeps the plus outside, (a + b), and flips to −ab in the middle. a³ − b³ keeps the minus outside, (a − b), and flips to +ab. Say it once before the exam and you will never mix up the four cube terms.